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I need to solve the following system: $$\begin{cases} A^TA=B &(1)\\ A\vec{x}=\vec{y} &(2)\\ \end{cases} $$ I need $A$, given $B$, $\vec{x}$ and $\vec{y}$.

$A$ and $B$ are both 3-by-3 matrices; $\vec{x}$ and $\vec{y}$ are 3-component vectors.

B is by construction a positive-definite matrix; $A$ is expected to be close to $I_3$.

In my problem, $B$ is actually the matrix defining the ellipsoid measured during the calibration procedure of a 3D sensor, and $A$ is the inverse of the sensitivity matrix to be found (it is not necessarily symmetric).

So we have 6 independent, quadratic equations in $(1)$ and 3 independent, linear equations in $(2)$ for 9 unknowns (the coefficients of $A$).

Does anyone have a clue about how to proceed to get an analytical solution?

Charlie
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2 Answers2

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Better than the QR decomposition I suggested above, the SVD decomposition might work. In fact, given $$ {\bf A} = {\bf U}\,{\bf \Sigma }\;\overline {\bf V} $$ the system above becomes $$ \left\{ \matrix{ \overline {\bf A} \;{\bf A} = {\bf V}\,{\bf \Sigma }\;\overline {\bf U} {\bf U}\,{\bf \Sigma }\;\overline {\bf V} = {\bf V}\,{\bf \Sigma }^{\bf 2} \,\overline {\bf V} = {\bf B}\quad \Leftrightarrow \quad {\bf B}\;{\rm diagonalizable} \hfill \cr {\bf U}\,{\bf \Sigma }\;\overline {\bf V} \,{\bf x} = {\bf y}\quad \Leftrightarrow \quad \overline {\bf x} \;{\bf V}\,{\bf \Sigma }^{\bf 2} \,\overline {\bf V} \,{\bf x} = \overline {\bf x} \;{\bf B}\,{\bf x} = \overline {\bf y} \;{\bf y} \hfill \cr} \right. $$ so that B,x,y are not totally independent (for real solutions to exist).
Therefrom ${\bf \Sigma }$ and ${\bf V}$ can be obtained from the normalized eigenvalue decomposition of ${\bf B}$.
${\bf U}$ is the rotation matrix that brings ${\bf \Sigma }\;\overline {\bf V} \,{\bf x}$ in ${\bf y}$.

G Cab
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    How do you compute the $\bf U$ that satisfies the equations in the original system?. The rotation matrix that brings $\bf \Sigma,\overline V,x$ into $\bf y$ is not unique. – Iban Cereijo Apr 09 '16 at 11:00
  • Indeed, more specifically, $U$ is an orthogonal matrix but does not have to be a rotation matrix. In other words, it belongs to $O(3)$ but not necessarily to $SO(3)$. – Charlie Apr 09 '16 at 12:41
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    Ok, got it, all of the solutions computed in this way satisfy the original equations, $\bf U$ can be any orthogonal matrix that takes $\bf \Sigma,V^T,x$ into $\bf y$. Great answer. – Iban Cereijo Apr 09 '16 at 12:51
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    True. I realise my problem is actually underdetermined... – Charlie Apr 09 '16 at 15:18
  • U orthogonal in general. But the physical nature of the application might suggest to exclude reflections (or maybe not if the sign of the measure is also under question, ...) – G Cab Apr 09 '16 at 16:07
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Hint

If you multiply the second equation by $A^T$ you get $A^T y = Bx$. One more multiplication by $A$ yields $A Bx = B y$. Now you can solve 9 by 9 linear sistem and check if the quadratic part is satisfied.

user26977
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